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irakobra [83]
3 years ago
13

The concentration of phenylephrine hydrochloride in a prefilled syringe is 10 mg/mL. What is the percentage concentration (w/v)

of phenylephrine in the injection. (Answer must be numeric; no units or commas; round the final answer to the nearest WHOLE number.)
Chemistry
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

The percentage concentration (w/v) of phenylephrine in the injection is 1 %.

Explanation:

The percentage concentration in w/v of the phenylephrine hydrochloride can be calculated as follows:

\%^{w}/_{v} = \frac{m}{V} \times 100

Where:

m: is the mass of the solute in grams

V: is the volume of the solution in milliliters

The concentration of the phenylephrine hydrochloride is 10 mg/mL, so the percentage concentration is:

\%^{w}/_{v} = \frac{m}{V} \times 100

\%^{w}/_{v} = \frac{10 \cdot 10^{-3} g}{1 mL} \times 100

\%^{w}/_{v} = 1 \%

Therefore, the percentage concentration (w/v) of phenylephrine in the injection is 1 %.

I hope it helps you!

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
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Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

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Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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